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Date: Tue, 8 Aug 2006 10:53:40 +1000
From: "Darren Jenkins" <darrenrjenkins@...il.com>
To: "Pavel Machek" <pavel@...e.cz>
Cc: torvalds@...l.org, "Zed 0xff" <zed.0xff@...il.com>,
kernel-janitors@...l.org, linux-kernel@...r.kernel.org
Subject: Re: [KJ] [patch] fix common mistake in polling loops
G'day
On 8/8/06, Pavel Machek <pavel@...e.cz> wrote:
> Hi!
>
> > >> Well, whoever wrote thi has some serious problems (in attitude
> > >> department). *Any* loop you design may take half a minute under
> > >> streange circumstances.
> >
> > 6.
> > common mistake in polling loops [from Linus]:
>
> Yes, Linus was wrong here. Or more precisely, he's right original code
> is broken, but his suggested "fix" is worse than the original.
>
> unsigned long timeout = jiffies + HZ/2;
> for (;;) {
> if (ready())
> return 0;
> [IMAGINE HALF A SECOND DELAY HERE]
> if (time_after(timeout, jiffies))
> break;
> msleep(10);
> }
>
> Oops.
>
> > >Actually it may be broken, depending on use. In some cases this loop
> > >may want to poll the hardware 50 times, 10msec appart... and your loop
> > >can poll it only once in extreme conditions.
> > >
> > >Actually your loop is totally broken, and may poll only once (without
> > >any delay) and then directly timeout :-P -- that will break _any_
> > >user.
> >
> > The Idea is that we are checking some event in external hardware that
> > we know will complete in a given time (This time is not dependant on
> > system activity but is fixed). After that time if the event has not
> > happened we know something has borked.
>
> But you have to make sure YOU CHECK READY AFTER THE TIMEOUT. Linus'
> code does not do that.
> Pavel
Sorry I did not realise that was your problem with the code.
Would it help if we just explicitly added a
if (ready())
return 0;
after the loop, in the example code? so people wont miss adding
something like that in?
Darren J.
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